Thermodynamics - water vapour cycle

[michealyap]

by solving the equation 0.393 x + 0.001091 ( 1 − x ) = 0.3066. i got x is 0.779 which is mass fraction of vapor then i got 0.221 for saturated liquid so for specific internal energy, i used interpolation to get the internal energy of saturated vapour and liquid the 150 degree celcius. I got 631.40075 kj/kg for saturated liquid, then i got 2559.49kj/kg for saturated vapor. so by multiplying mass fraction of respected specific internal energy, 631.4075*= specific internal energy for saturated liquid in the system at point 2, then 2559.49kj/kg/for saturated vapor in the system at point 2.To get Ufg = 2558.49-631.4065=1928.081 kj/kg. =using the fomula Uavg= Uf+ X*Ufg = 631.4075 +0.78 *1928.089= 2135.310kj/kg. specific internal energy is 2135.310kj/kg at point 2.

 

[Chestermiller]

by solving the equation 0.393 x + 0.001091 ( 1 − x ) = 0.3066. i got x is 0.779 which is mass fraction of vapor then i got 0.221 for saturated liquid so for specific internal energy, i used interpolation to get the internal energy of saturated vapour and liquid the 150 degree celcius. I got 631.40075 kj/kg for saturated liquid, then i got 2559.49kj/kg for saturated vapor. so by multiplying mass fraction of respected specific internal energy, 631.4075*= specific internal energy for saturated liquid in the system at point 2, then 2559.49kj/kg/for saturated vapor in the system at point 2.To get Ufg = 2558.49-631.4065=1928.081 kj/kg. =using the fomula Uavg= Uf+ X*Ufg = 631.4075 +0.78 *1928.089= 2135.310kj/kg. specific internal energy is 2135.310kj/kg at point 2.

Your approach looks correct. But there was actually no need to do any interpolation. In your set of steam tables, there should not only be a table for saturated conditions at specified pressures, but also a table for saturated conditions at 10 degree increments of temperature. For 150 C, that table gives the following values:

Equilibrium pressure = 4.758 bars
Specific volume of liquid = 0.001091 m^3/kg
Specific volume of vapor = 0.393 M^3/kg
Internal energy of liquid = 631.7 kj/kg
Internal energy of vapor = 2559 kj/kg
Enthalpy of saturated liquid = 632.2 kj/kg
Enthalpy of saturated vapor = 2746 kj/kg

 

[michealyap]

hi, after that i do not know how to proceed. 1st law of thermodynamics change of internal energy = Q- W. for the process of 1 to 2, constant volume so work done on this is equal to zero. therefore internal energy of procee 1 - process 2 = 2597.3 kj/kg - 2135.310 kj/kg = 462kj/kg. heat transfer out of the system. then how to find the work done for the system?

 

[Chestermiller]

hi, after that i do not know how to proceed. 1st law of thermodynamics change of internal energy = Q- W. for the process of 1 to 2, constant volume so work done on this is equal to zero. therefore internal energy of procee 1 - process 2 = 2597.3 kj/kg - 2135.310 kj/kg = 462kj/kg. heat transfer out of the system. then how to find the work done for the system?

Didn't you just say that the work is zero?

 

[michealyap]

Hello, yea i thought the question is asking the whole system first is constant volume , then is condensed isothermally. so the specific workdone is on the process of condensation.

 

[Chestermiller]

OK. Let's talk about the process step between points 2 and 3. What can you tell me about that step (in terms of the first law)? What are the conditions at the end of that step?

 

[michealyap]

for process between point 2 and 3, 1st law of thermodynamics , the change of internal energy is equal to zero. so Q=W.

 

[Chestermiller]

for process between point 2 and 3, 1st law of thermodynamics , the change of internal energy is equal to zero. so Q=W.

This is definitely incorrect. You are condensing the vapor to a liquid at constant temperature and pressure. Did you think that the internal energy is a function only of temperature even if there is a phase change? Start with the equation $$\Delta U=Q-P\Delta V$$What does this give you for Q?

 

[michealyap]

Q=heat transfer out. how do we know that it is condensed at constant pressure. In the same time, how do we unable to find the internal energy at point 3, which only contain water.

 

[Chestermiller]

Q=heat transfer out. how do we know that it is condensed at constant pressure. In the same time, how do we unable to find the internal energy at point 3, which only contain water.

If it is condensed at constant temperature (problem statement), it must be at the equilibrium vapor pressure at that temperature. So the pressure is constant.

Regarding your question about the internal energy at point 3, you sure can find the internal energy at point 3 from the saturated steam table. You can also determine the enthalpy at points 2 and 3, which may be easier to work with.

Don't forget to use the saturated steam table at specified temperatures, not specified pressures, to avoid having to interpolate.

 

[michealyap]

ok.. since at point 3 all the water vapor has condensed to become water. the specific internal energy is 631.68 kj/kg for saturated liquid. pressure is 4.758bar. what is the delta V at here? as the formula U2-U3 = Q -PdeltaV .

Specific internal energy = 2135.7 kj/kg at point 2
Specific internal energy = 631.68 kj/kg at point 3

 

[Chestermiller]

ok.. since at point 3 all the water vapor has condensed to become water. the specific internal energy is 631.68 kj/kg for saturated liquid. pressure is 4.758bar. what is the delta V at here? as the formula U2-U3 = Q -PdeltaV .

Specific internal energy = 2135.7 kj/kg at point 2
Specific internal energy = 631.68 kj/kg at point 3

Well, you know that the specific volume at point 2 (averaged over the mass of vapor and liquid) is 0.3066 m^3/kg. And you know that, at point 3, it is all saturated liquid at 0.001091 m^3/kg.

A second way of doing all this is to recognize that, for every kg of water in the container, you have 0.779 kg of vapor at point 2. So, the heat removal to condense this is the heat of vaporization (listed in the table) times 0.779.

A third way of doing this is to get the enthalpy per kg of mixture at point 2. This is done the same way you got the internal energy at point 2. From this, you subtract the enthalpy of the saturated liquid at point 3 to get the heat required to be removed per kg.

See if you can show that all three of these methods give exactly the same results for the heat Q.

 

[michealyap]

it did not work so well for the method 1. change of U = Q - Pdelta V . so, 2135.7 kj/kg - 631.68/kj/kg = Q - 4.758 *10^5 * (0.3066-0.001091(m^3/kg)) , therefore .143.857kj/kg = Q , this the Q i obtained..

For second method how to get the heat of vaporiztion?

 

[Chestermiller]

it did not work so well for the method 1. change of U = Q - Pdelta V . so, 2135.7 kj/kg - 631.68/kj/kg = Q - 4.758 *10^5 * (0.3066-0.001091(m^3/kg)) , therefore .143.857kj/kg = Q , this the Q i obtained..

For second method how to get the heat of vaporiztion?

To get the work, you need to subtract the initial volume from the final volume. And, to get the change in internal energy, you need to subtract the initial internal energy from the final internal energy. So, $$P\Delta V=4.758\times 10^5(0.001091-0.3066)=-145000\ J/kg=-145\ kJ/kg$$. And, $$\Delta U=631.68-2135.7=-1504\ kJ/kg=Q-(-145)=Q+145$$ and so $$Q=-1649\ kJ/kg$$

For the 2nd method, to get the heat of vaporization, you subtract the specific enthalpy of the saturated liquid from the specific enthalpy of the saturated vapor (in the table).

 

[michealyap]

erm.. that is awesome. thank you.. really have a clearer picture.